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Question: In a series circuit \(R = 300\Omega,L = 0.9H\), \(C = 2.0\mu F\)and \(\omega = 1000rad/sec.\) The im...

In a series circuit R=300Ω,L=0.9HR = 300\Omega,L = 0.9H, C=2.0μFC = 2.0\mu Fand ω=1000rad/sec.\omega = 1000rad/sec. The impedance of the circuit is

A

1300 Ω

B

900 Ω

C

500 Ω

D

400 Ω

Answer

500 Ω

Explanation

Solution

Z=R2+(ωL−1ωC)2=(300)2+(1000×0.9−11000×2×10−6)2Z = \sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}} = \sqrt{(300)^{2} + \left( 1000 \times 0.9 - \frac{1}{1000 \times 2 \times 10^{- 6}} \right)^{2}}

⇒ Z=(300)2+(400)2=500ΩZ = \sqrt{(300)^{2} + (400)^{2}} = 500\Omega.